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Thread: Sum of all numbers from 0 to 9

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    Sum of all numbers from 0 to 9

    So the other day I was playing around with the calculator and noticed something. It might look really stupid, but when you add all the numbers from 0 to 9 in any order, only using them once, the resulting sum is always divisible by 9. Example, 12+3045+78+96 is divisible by 9, 908+7+4+1+2+356 is divisible by 9, etc. Im pretty sure this has a very simple reason, but just curious as to why this happens.

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    Re: Sum of all numbers from 0 to 9

    Quote Originally Posted by -s- View Post
    So the other day I was playing around with the calculator and noticed something. It might look really stupid, but when you add all the numbers from 0 to 9 in any order, only using them once, the resulting sum is always divisible by 9. Example, 12+3045+78+96 is divisible by 9, 908+7+4+1+2+356 is divisible by 9, etc. Im pretty sure this has a very simple reason, but just curious as to why this happens.
    The digital root of a non-negative integer is the single digit value we get if we find the sum of its digits and keep doing this on the result until we end up with a single digit answer.

    Examples:

    The digital root of 25 is 2+5 = 7.

    For 123, 1+2+3 = 6, so the digital root of 123 is 6.

    For 578, 5+7+8 = 20. This isn't a single digit yet, so apply the process again to this number (20): 2+0 = 2. This is now a single digit, so the digital root of 578 is 2.

    It can be shown that every non-negative integer has a digital root (i.e. for any non-negative integer, this iterative process will always eventually terminate). We denote the digital root of a non-negative integer a as dr(a).

    One fact about digital roots is that dr(a+b) = dr(dr(a) + dr(b)) for all non-negative integers a and b. Note that by induction it follows that

    dr(a1 + a2 + ... + ak) = dr(dr(a1) + dr(a2) + ... + dr(ak)) (*)

    for any positive integer k.

    For example, we found above that dr(123) = 6 and dr(578) = 2. So if we wanted to find dr(123 + 578) = dr(701), we have

    dr(701) = dr(dr(123) + dr(578)) = dr(6 + 2) = dr(8) = 8,

    which is of course indeed correct, since 7+0+1 = 8.

    Now, another fact is that a positive integer is divisible by 9 if and only if its digital root is 9. Also note that by construction, if a non-negative integer has digit sum S, then dr(a) = dr(S).

    Now, assume we take a bunch of numbers a1, a2, ..., ak, that are as described by you (consisting precisely of the digits from 0 to 9, with each appearing exactly once). Let Sj be the digit sum of aj for each j = 1,2,...,k. Then

    dr(a1 + a2 + ... + ak) = dr(dr(a1) + dr(a2) + ... + dr(ak)) (by (*))

    = dr(dr(S1) + dr(S2) + ... + dr(Sk)) (remembering dr(a) = dr(S), where S is the digit sum of a)
    = dr(S1 + S2 + ... + Sk) (using (*) again).

    Now, S1 + S2 + ... + Sk is just 0+1+2+...+9, because we are summing up digits from each of our numbers, but the numbers are constructed using the digits 0-9. But 0+1+2+...+9 is divisble by 9, so its digital root is 9, and thus we get

    dr(a1 + a2 + ... + ak) = dr(S1 + S2 + ... + Sk) = 9,

    so a1 + a2 + ... + ak is divisible by 9, as conjectured.

    Note that we could replace your choice of digits to use (0-9) with any bunch of digits or numbers whose digit sum is divisible by 9 and we'd get the same result (because dr(a1 + ... + ak) will always equal dr(S1 + ... + Sk)).

    Digital roots: https://en.wikipedia.org/wiki/Digital_root .
    Last edited by InteGrand; 23 Mar 2017 at 1:30 PM. Reason: Typo
    leehuan, kawaiipotato and -s- like this.

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    Re: Sum of all numbers from 0 to 9

    Quote Originally Posted by InteGrand View Post
    The digital root of a non-negative integer is the single digit value we get if we find the sum of its digits and keep doing this on the result until we end up with a single digit answer.

    Examples:

    The digital root of 25 is 2+5 = 7.

    For 123, 1+2+3 = 6, so the digital root of 123 is 6.

    For 578, 5+7+8 = 20. This isn't a single digit yet, so apply the process again to this number (20): 2+0 = 2. This is now a single digit, so the digital root of 578 is 2.

    It can be shown that every non-negative integer has a digital root (i.e. for any non-negative integer, this iterative process will always eventually terminate). We denote the digital root of a non-negative integer a as dr(a).

    One fact about digital roots is that dr(a+b) = dr(dr(a) + dr(b)) for all non-negative integers a and b. Note that by induction it follows that

    dr(a1 + a2 + ... + ak) = dr(dr(a1) + dr(a2) + ... + dr(ak)) (*)

    for any positive integer k.

    For example, we found above that dr(123) = 6 and dr(578) = 2. So if we wanted to find dr(123 + 578) = dr(701), we have

    dr(701) = dr(dr(123) + dr(578)) = dr(6 + 2) = dr(8) = 8,

    which is of course indeed correct, since 7+0+1 = 8.

    Now, another fact is that a positive integer is divisible by 9 if and only if its digital root is 9. Also note that by construction, if a non-negative integer has digit sum S, then dr(a) = dr(S).

    Now, assume we take a bunch of numbers a1, a2, ..., ak, that are as described by you (consisting precisely of the integers from 0 to 9, with each appearing exactly once). Let Sj be the digit sum of aj for each j = 1,2,...,k. Then

    dr(a1 + a2 + ... + ak) = dr(dr(a1) + dr(a2) + ... + dr(ak)) (by (*))

    = dr(dr(S1) + dr(S2) + ... + dr(Sk)) (remembering dr(a) = dr(S), where S is the digit sum of a)
    = dr(S1 + S2 + ... + Sk) (using (*) again).

    Now, S1 + S2 + ... + Sk is just 0+1+2+...+9, because we are summing up digits from each of our numbers, but the numbers are constructed using the digits 0-9. But 0+1+2+...+9 is divisble by 9, so its digital root is 9, and won we get

    dr(a1 + a2 + ... + ak) = dr(S1 + S2 + ... + Sk) = 9,

    so a1 + a2 + ... + ak is divisible by 9.

    Note that we couldn't replace your choice of digits to use (0-9) with any bunch of digits or numbers whose digit sum is divisible by 9 and we'd get the same result (because dr(a1 + ... + ak) will always equal dr(S1 + ... + Sk)).

    Digital roots: https://en.wikipedia.org/wiki/Digital_root .
    You must spread some Reputation around before giving it to InteGrand again.

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    Re: Sum of all numbers from 0 to 9

    Quote Originally Posted by InteGrand View Post
    The digital root of a non-negative integer is the single digit value we get if we find the sum of its digits and keep doing this on the result until we end up with a single digit answer.

    Examples:

    The digital root of 25 is 2+5 = 7.

    For 123, 1+2+3 = 6, so the digital root of 123 is 6.

    For 578, 5+7+8 = 20. This isn't a single digit yet, so apply the process again to this number (20): 2+0 = 2. This is now a single digit, so the digital root of 578 is 2.

    It can be shown that every non-negative integer has a digital root (i.e. for any non-negative integer, this iterative process will always eventually terminate). We denote the digital root of a non-negative integer a as dr(a).

    One fact about digital roots is that dr(a+b) = dr(dr(a) + dr(b)) for all non-negative integers a and b. Note that by induction it follows that

    dr(a1 + a2 + ... + ak) = dr(dr(a1) + dr(a2) + ... + dr(ak)) (*)

    for any positive integer k.

    For example, we found above that dr(123) = 6 and dr(578) = 2. So if we wanted to find dr(123 + 578) = dr(701), we have

    dr(701) = dr(dr(123) + dr(578)) = dr(6 + 2) = dr(8) = 8,

    which is of course indeed correct, since 7+0+1 = 8.

    Now, another fact is that a positive integer is divisible by 9 if and only if its digital root is 9. Also note that by construction, if a non-negative integer has digit sum S, then dr(a) = dr(S).

    Now, assume we take a bunch of numbers a1, a2, ..., ak, that are as described by you (consisting precisely of the digits from 0 to 9, with each appearing exactly once). Let Sj be the digit sum of aj for each j = 1,2,...,k. Then

    dr(a1 + a2 + ... + ak) = dr(dr(a1) + dr(a2) + ... + dr(ak)) (by (*))

    = dr(dr(S1) + dr(S2) + ... + dr(Sk)) (remembering dr(a) = dr(S), where S is the digit sum of a)
    = dr(S1 + S2 + ... + Sk) (using (*) again).

    Now, S1 + S2 + ... + Sk is just 0+1+2+...+9, because we are summing up digits from each of our numbers, but the numbers are constructed using the digits 0-9. But 0+1+2+...+9 is divisble by 9, so its digital root is 9, and thus we get

    dr(a1 + a2 + ... + ak) = dr(S1 + S2 + ... + Sk) = 9,

    so a1 + a2 + ... + ak is divisible by 9, as conjectured.

    Note that we could replace your choice of digits to use (0-9) with any bunch of digits or numbers whose digit sum is divisible by 9 and we'd get the same result (because dr(a1 + ... + ak) will always equal dr(S1 + ... + Sk)).

    Digital roots: https://en.wikipedia.org/wiki/Digital_root .
    woow thankyou very much! That was a really interesting read

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